Question

The direction angles of the line $x=4z+3,y=2-3z$ are $\alpha ,\beta$ and $\gamma$, then $\cos \alpha +\cos \beta +\cos \gamma =$ ________.

• A. $\frac{2}{\sqrt{26}}$
• B. $\frac{8}{\sqrt{26}}$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{2}{\sqrt{26}}$

$\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z}{1}$

$\cos \alpha$is given by $\cos \alpha =\frac{a}{\sqrt{a^2+b^2+c^2}}$

$\cos \beta =\frac{b}{\sqrt{a^2+b^2+c^2}}$

$\cos \gamma =\frac{c}{\sqrt{a^2+b^2+c^2}}$

$\cos \alpha =\frac{4}{\sqrt{9+16+1}}=\frac{4}{\sqrt{26}}$

$\cos \beta =\frac{4}{\sqrt{9+16+1}}=\frac{-3}{\sqrt{26}}$

$\cos \gamma =\frac{1}{\sqrt{9+16+1}}=\frac{1}{\sqrt{26}}$

$\cos \alpha =\frac{4}{\sqrt{26}},\cos \beta =\frac{-3}{\sqrt{26}},\cos \gamma =\frac{1}{\sqrt{26}}$

$\therefore \cos \alpha +\cos \beta +\cos \gamma =\frac{2}{\sqrt{26}}$.